11th Class Physics Physical World / भौतिक जगत Question Bank 11th CBSE Physics Mathematical Tools, Units & Dimensions

  • question_answer
    Check the correctness of the relation \[c=\frac{1}{\sqrt{{{\mu }_{0}}{{\in }_{0}}}}\]where the symbols have their usual meaning.

    Answer:

                    c = velocity of light in vacuum =                                      \[~\left[ {{\text{M}}^{0}}{{\text{L}}^{\text{1}}}{{\text{T}}^{-\text{1}}} \right]\]                                                                                                     \[{{\mu }_{0}}\] = absolute magnetic permeability of free space =                                           \[~\left[ \text{ML}{{\text{T}}^{-\text{2}}}{{\text{A}}^{-\text{2}}} \right]\]                                                                                                         \[{{\in }_{0}}\]  = absolute electrical permittivity of free space =         \[\left[ {{\text{M}}^{-\text{1}}}{{\text{L}}^{-\text{3}}}{{\text{T}}^{\text{4}}}{{\text{A}}^{\text{2}}} \right]\] . Now, L.H.S. =                       \[\text{c}=\text{ }\left[ {{\text{M}}^{0}}{{\text{L}}^{\text{1}}}{{\text{T}}^{-\text{1}}} \right]\] R.H.S. = \[\frac{1}{\sqrt{{{\mu }_{0}}{{\in }_{0}}}}=\frac{1}{\sqrt{[ML{{T}^{-2}}{{A}^{-2}}][{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}]}}=\frac{1}{\sqrt{{{M}^{0}}{{L}^{-2}}{{T}^{2}}}}=\text{ }\left[ {{\text{M}}^{0}}{{\text{L}}^{\text{1}}}{{\text{T}}^{-\text{1}}} \right]\] As L.H.S. = R.H.S., dimensionally.                                                                                                       \[\therefore\] formula is correct.


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